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A charge q is placed at the centre of the cubical vessel (with one face open) as shown in figure. The flux of the electric field through the surface of the vessel is

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$\frac{q}{ε_{0}}$

$\frac{5 q}{6 ε_{0}}$

$\frac{q}{4 ε_{0}}$

answer is D.

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Detailed Solution

Total flux through closed cubical vessel = $\frac{q}{ε_{0}}$ .
& Flux through one face = $\frac{1}{6} (\frac{q}{ε_{0}})$
So, total flux passing through given cubical vessel is = $5 (\frac{q}{6 ε_{0}})$ ;

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